similarly if is a constant vector, then:
also, with a little bit more effort, you can show that for any constant matrix we have:
now solving the problem is very easy:
now in (4): the derivative of is clearly 0. by (1) the derivative of is by (2) the derivative of is finally by (3) the derivative
of is thus:
recall that if where are scalar functions of then where is a defined by using this youare you sure it's not instead ?
can very easily show that if is a constant matrix, then
thus by (5) and (6):